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I can do all of this question except the last part: Let $N$ be a normal subgroup of a finite group $G$ of prime index $p$. Show that if $H$ is a subgroup of $G$ then $H\cap N$ is a normal subgroup of $H$ of index $1$ or $p$. Could I have some help please?

I've shown that $H\cap N$ is a subgroup using the one-step test, and I've shown it's normal by conjugating it with an element of $H$.

I'm stuck on identifying the index of $H\cap N$. I'm guessing it involves Lagrange's Theorem, and I tried to come up with a homomorphism from $H$ to something with kernel $H\cap N$, but with no success.

I'd prefer hints to a full solution, if that's alright.

Thank you very much for all your help.

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  • $\begingroup$ Have you heard of the second isomorphism theorem? $\endgroup$ – ArtW Jan 16 '18 at 17:10
  • $\begingroup$ Heard of it, but haven't studied it yet, and I would rather solve this with methods I am familiar with. $\endgroup$ – John Smith Jan 16 '18 at 17:11
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Even without the 2nd isomorphism theorem, it could be you've already studied that

$$[H:H\cap N]=[NH:N]$$

and since the last one divides the index of $\;N\;$ in G we're done.

To prove the above equality you could try the following: define

$$f:H\to NH/N\;,\;\;f(h):=hN$$

(1) Show $\;f\;$ is a homomorphism

(2) Show $\;H\cap N=\ker f\;$

(3) Show that $\;f\;$ is surjective

(4) and finally use the first isomorphism theorem ...

In fact, the above is almost-almost the proof of the 2nd isomorphism theorem...:)

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